If you are right that 2 is an eigenvalue, then there must exist a vector, , x, y, z not all 0, such that .

That gives the three equations -2x-8 y- 12z= 2x, x+ 4y+ 4z= 2y, and z= 2z.
Of course, that last equations tells us that z= 0 so the first two become -2x- 8y= 2x, or 8y= -4x so that x= -2y, and x+ 4y= 2y and, again, x= -2y. The fact that there exist an infinite number of solutions to those equations confirms that 2 is an eigenvalue. Any eigenvector corresponding to eigenvalue 2 is of the form .

If you are right that 2 is an eigenvalue, then there must exist a vector, , x, y, z not all 0, such that .

That gives the three equations -2x-8 y- 12z= 2x, x+ 4y+ 4z= 2y, and z= 2z.
Of course, that last equations tells us that z= 0 so the first two become -2x- 8y= 2x, or 8y= -4x so that x= -2y, and x+ 4y= 2y and, again, x= -2y. The fact that there exist an infinite number of solutions to those equations confirms that 2 is an eigenvalue. Any eigenvector corresponding to eigenvalue 2 is of the form .